/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) UsableRulesProof [EQUIVALENT, 0 ms]
        (12) QDP
        (13) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (14) YES
    (15) QDP
        (16) QDPOrderProof [EQUIVALENT, 0 ms]
        (17) QDP
        (18) PisEmptyProof [EQUIVALENT, 0 ms]
        (19) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(minus(x, y), double(y)))

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS(s(x), s(y)) -> MINUS(x, y)
   DOUBLE(s(x)) -> DOUBLE(x)
   PLUS(s(x), y) -> PLUS(x, y)
   PLUS(s(x), y) -> PLUS(x, s(y))
   PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
   PLUS(s(x), y) -> MINUS(x, y)
   PLUS(s(x), y) -> DOUBLE(y)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(minus(x, y), double(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 2 less nodes.
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(4)
Complex Obligation (AND)

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(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DOUBLE(s(x)) -> DOUBLE(x)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(minus(x, y), double(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   DOUBLE(s(x)) -> DOUBLE(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(8) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*DOUBLE(s(x)) -> DOUBLE(x)
The graph contains the following edges 1 > 1


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(9)
YES

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(10)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS(s(x), s(y)) -> MINUS(x, y)

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(minus(x, y), double(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(11) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   MINUS(s(x), s(y)) -> MINUS(x, y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(13) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*MINUS(s(x), s(y)) -> MINUS(x, y)
The graph contains the following edges 1 > 1, 2 > 2


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(14)
YES

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(15)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   PLUS(s(x), y) -> PLUS(x, s(y))
   PLUS(s(x), y) -> PLUS(x, y)
   PLUS(s(x), y) -> PLUS(minus(x, y), double(y))

The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(minus(x, y), double(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(16) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   PLUS(s(x), y) -> PLUS(x, s(y))
   PLUS(s(x), y) -> PLUS(x, y)
   PLUS(s(x), y) -> PLUS(minus(x, y), double(y))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( PLUS_2(x_1, x_2) ) = max{0, 2x_1 - 1}
POL( minus_2(x_1, x_2) ) = x_1
POL( 0 ) = 2
POL( s_1(x_1) ) = x_1 + 2
POL( double_1(x_1) ) = x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)


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(17)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   minus(x, 0) -> x
   minus(s(x), s(y)) -> minus(x, y)
   double(0) -> 0
   double(s(x)) -> s(s(double(x)))
   plus(0, y) -> y
   plus(s(x), y) -> s(plus(x, y))
   plus(s(x), y) -> plus(x, s(y))
   plus(s(x), y) -> s(plus(minus(x, y), double(y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(18) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
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(19)
YES